Given inorder and postorder traversal of a tree, construct the binary tree.

 Notice

You may assume that duplicates do not exist in the tree.

 
Example

Given inorder [1,2,3] and postorder [1,3,2], return a tree:

  2
 / \
1   3
分析：
这是非常典型的递归问题，postorder的最后一个是root，在inorder里面找出root的位置，左边部分为左子树，右边部分为右子树，稍微麻烦的部分就是确定左子树和右子树的starting index。
 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode buildTree(int[] inorder, int[] postorder) {
12         if (inorder == null || postorder == null || postorder.length == 0 || inorder.length != postorder.length) return null;
13         return buildTree(inorder, 0, postorder, 0, inorder.length);
14     }
15     
16     public TreeNode buildTree(int[] inorder, int inStart, int[] postorder, int postStart, int length) {
17         if (length == 0) return null;
18         TreeNode root = new TreeNode(postorder[postStart + length - 1]);
19         int index = findIndex(inorder, root.val);
20         root.left = buildTree(inorder, inStart, postorder, postStart, index - inStart);
21         root.right = buildTree(inorder, index + 1, postorder, postStart + (index - inStart), length - (index - inStart) - 1);
22         return root;    
23     }
25     public int findIndex(int[] inorder, int val) {
26         for (int i = 0; i < inorder.length; i++) {
27             if (inorder[i] == val) {
28                 return i;
29             }
30         }
31         return -1;
32     }
33 }